1 / 40

5.6 Laws of Logarithms 5.7 Exponential Equations; Changing Base

5.6 Laws of Logarithms 5.7 Exponential Equations; Changing Base. Objectives : 1. Compare & recall the properties of exponents 2. Deduce the properties of logarithms from/by comparing the properties of exponents Use the properties of logarithms Solve the exponential equations.

kasa
Download Presentation

5.6 Laws of Logarithms 5.7 Exponential Equations; Changing Base

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 5.6 Laws of Logarithms5.7 Exponential Equations; Changing Base Objectives: 1. Compare & recall the properties of exponents 2. Deduce the properties of logarithms from/by comparing the properties of exponents • Use the properties of logarithms • Solve the exponential equations

  2. Since the logarithmic function y = logb x is the inverse of the exponential function y = bx, the laws of logarithms are very closely related to the laws of exponent. Pre-Knowledge For any b, c, u, v +, and b≠ 1, c ≠ 1, there exists some x, y , such that u = bx, v = by By the previous section knowledge, as long as taking x = logbu, y = logbv

  3. 1. Product of Power am an = am+n Pre-Knowledge For any b, c, u, v +, and b≠ 1, c ≠ 1, there exists some x, y , such that u = bx, v = by By the previous section knowledge, as long as taking x = logbu, y = logbv 1. Product Property logbuv = logbu + logbv Proof logbuv = logb(bxby)= logbb x+y = x + y = logbu + logbv

  4. 2. Quotient of Power Pre-Knowledge For any b, c, u, v +, and b≠ 1, c ≠ 1, there exists some x, y , such that u = bx, v = by By the previous section knowledge, as long as taking x = logbu, y = logbv 2. Quotient Property Proof

  5. Equal Power • am = an iff m = n • Equal Property • logbu = logbv iff u = v Proof logbu – logbv = 0

  6. 4. Power of Power (am)n = amn Pre-Knowledge For any b, c, u, v +, and b≠ 1, c ≠ 1, there exists some x, y , such that u = bx, v = by By the previous section knowledge, as long as taking x = logbu, y = logbv 4. Power Property logbuk = k logbu Proof logbuk = logb(bx)k= logbb kx = kx = k logbu

  7. 5. Change-of-Base Formula Proof Note that bx = u, logbu = x Taking the logarithm with base c at both sides: logcbx = logcu or x logcb = logcu

  8. 6. Reciprocal Formula Proof

  9. 7. Raise Power Formula Proof

  10. Example 1 Assume that log95 = a, log911 = b, evaluate • log9 (5/11) • log955 • log9125 • log9(121/45) • log9275

  11. Example 2 Expanding the expression • ln(3y4/x3) ln(3y4/x3) = ln(3y4)– lnx3 = ln3 + lny4 – lnx3 = ln3 + 4 ln|y| – 3 lnx b) log3125/6x9 log3125/6x9 = log3125/6 + log3x9 = 5/6 log312 + 9 log3x = 5/6 log3(3· 22) + 9 log3x = 5/6 (log33 + log322) + 9 log3x = 5/6 ( 1 + 2 log32) + 9 log3x

  12. Example 3 Condensing the expression a) 3 ( ln3 – lnx ) + ( lnx – ln9 ) 3 ( ln3 – lnx ) + ( lnx – ln9 ) = 3 ln3 – 3 lnx + lnx – 2 ln3 = ln3 – 2 lnx = ln(3/x2) b) 2 log37 – 5 log3 x + 6 log9 y2 2 log37 – 5 log3 x + 6 log9 y2 = log349 – log3 x5 + 6 ( log3 y2/ log39) = log3(49/x5) + 3 log3 y2 = log3(49y6/x5)

  13. Practice A) P. 199 Q 7 – 18 • P. 199 Q 19 – 20 How do you change to make it to be true? True or Falselog ay = – log 1/ay c) P. 200 Q 7 – 27 (odd)

  14. Example 4 Calculate log48 and log615 using common and natural logarithms. a) log48 log48 = log8 / log4 = 3 log2 / (2 log2) = 3/2 log48 = ln8 / ln4 = 3 ln2 / (2 ln2) = 3/2 b) log615 = log15 / log6 = 1.511

  15. in terms of Example 5. Express

  16. More on Expand/Condense logarithmic expressions Example 6 Expand

  17. Example 7 Expand in terms of sums and differences of logarithms More on Expand/Condense logarithmic expressions

  18. More on Expand/Condense logarithmic expressions Example 8 Expand to express all powers as factors

  19. Example 9 Condense to a single logarithm. More on Expand/Condense logarithmic expressions

  20. Assignment: 5.6 P. 196 #36 – 44 (even) P. 200 #2 – 22 (even), 21 – 33 (odd), 41, 43, 45

  21. Solving Exponential Equations • One way to solve exponential equations is to use the property that if 2 powers with the same base are equal, then their exponents are equal. • For b > 0 and b≠1 if bx = by, then x = y

  22. Solve by Equating Exponents Example 10: Solve 43x = 8x+1 (22)3x = (23)x+1 rewrite with same base 26x = 23x+3 6x = 3x + 3 x = 1 Check → 43*1 = 81+1 64 = 64

  23. Your turn! Solve: 16x = 32x–1 24x = 32x–1 24x = (25)x–1 4x = 5x – 5 x = 5 Be sure to check your answer!!!

  24. When you can’t rewrite using the same base, you can solve by taking a log of both sides or use the definition of log Example 11: Solve 2x = 7 log22x = log27 x = log27 x = ≈ 2.807

  25. Example 12: Solve 102x – 3 + 4 = 21 When you can’t rewrite using the same base, you can solve by taking a log of both sides or use the definition of log 102x – 3 = 17 log10102x – 3 = log1017 2x – 3 = log 17 2x = 3 + log17 x = ½(3 + log17) ≈ 2.115

  26. Your turn! Solve: 5x + 2 + 3 = 25 5x+2 = 22 log55x+2 = log522 x + 2 = log522 x = (log522) – 2 = (log22/log5) – 2 ≈ –0.079

  27. More on Solving Exponential Equations Example 13: Solve ex – 3e-x = 2 [Answer] Multiply ex at both sides of the equation: ex(ex – 3e-x ) = 2ex e2x–3 = 2ex e2x–2ex –3 = 0 (ex)2–2(ex)–3 = 0 Denote ex= u, then (u)2–2(u)–3 = 0 (u)2–2(u)–3 = 0 (u – 3)(u+ 1) = 0 u = 3, or u= –1 ex= 3, or ex = –1 (discard) x = ln3

  28. Solving Logarithmic Equations 1. To solve use the property for logs with the same base: • b, x, y+ and b  1 • If logb x = logb y, then x = y 2. When you can’t rewrite both sides as logs with the same base exponentiate each side • b, x+ and b  1 • if logb x = y, then x = by • This can get the expression in the log out of the log simply.

  29. Solving Logarithmic Equations Example 14: Newton’s Law of Cooling • The temperature T of a cooling substance at time t (in minutes) is: • T = (T0 – TR) e-rt + TR • T0= initial temperature • TR= room temperature • r = constant cooling rate of the substance

  30. Solving Logarithmic Equations Example 14: You’re cooking stew. When you take it off the stove the temp. is 212°F. The room temp. is 70°F and the cooling rate of the stew is r = 0.046. How long will it take to cool the stew to a serving temp. of 100°? T = (T0 – TR) e -rt + TR T0 = 212, TR = 70, T = 100 r = 0.046 So solve: 100 = (212 – 70)e -0.046t + 70

  31. Solving Logarithmic Equations 30 = 142e -0.046t(subtract 70) 15/71 = e -0.046t(divide by 142) • How do you get the variable out of the exponent? ln(15/71) = lne-.046t (take the ln of both sides) ln(15/71) = – 0.046t ln(15/71)/(– 0.046) = t t =(ln15 – ln71)/(– 0.046) = t ≈ – 1.556/(– 0.046) t ≈ 33.8 about 34 minutes to cool!

  32. Example 15: Solve log3(5x – 1) = log3(x + 7) 5x – 1 = x + 7 4x = 8 x = 2 and check log3(52 – 1) = log3(2 + 7) log39 = log39 Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions. Solving Logarithmic Equations

  33. Example 16: Solve log5(3x + 1) = –2 3x + 1 = 5-2 3x + 1 = 1/25 x = –8/25 and check Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions. Solving Logarithmic Equations

  34. More on Solving Logarithmic Equations Example 16: Solve log5x + log(x + 1) = 2 log[5x(x + 1)]= 2 (product property) log (5x2 + 5x) = 2 5x2 + 5x = 100 x2 + x – 20 = 0 (subtract 100 and divide by 5) (x + 5)(x – 4) = 0 x = – 5, or x = 4 check and you’ll see x = 4 is the only solution.

  35. Your Turn! Solve log2x + log2(x – 7) = 3 log2 [x(x – 7)]= 3 log2 (x2 – 7x) = 3 x2 – 7x = 23 x2 – 7x = 8 x2 – 7x – 8 = 0 (x – 8)(x + 1) = 0 x = 8 x = –1 Check log28 + log2(8 – 7) =3 3 + 0 = 3

  36. One More!Solve log6(x – 2) + log6(x + 3) = 2 log6(x – 2) + log6(x + 3) = 2 log6 [(x – 2)(x + 3)] = 2 log6 (x2 + x – 6) = 2 x2 + x – 6 = 36 x2 + x – 42 = 0 (x – 6)(x + 7)=0 x = 6 x = –7 Check log64 + log69 =2 log636 = 2 Check log64 + log69 =2 log636 = 2

  37. Challenge!Example 17: Solve log2x + log4(x2 – 4x + 4) = 3 log4 x2 + log4(x – 2)2 = 3 (Raise Power Formula) log4 [x2(x – 2)2] = 3 x2(x – 2)2 = 43 [x(x – 2)]2 = 64 A2B2 = (AB)2 x(x – 2) = ±8 A2 = 64, A = ±8 x2 – 2x – 8 = 0 or x2 – 2x + 8 =0 x = 4 x = –2 No real solution

  38. Challenge!Solve log2x + ½ log2(x – 1) = 1 2log2x + log2(x – 1) = 2 log2x2 + log2(x - 1) = 2 log2 [x2(x – 1)] = 2 x2(x – 1) = 4 x3– x2 – 4 = 0 (Rational Zero Theorem) (x – 2)(x2 + x + 2) = 0 x – 2 = 0 or x2 + x + 2 = 0 x = 2 No real solution Check log22 + ½ log21 =1 2 + 0 = 2

  39. Challenge Simplify (No calculator) 1) 2) 3) 4) 5) Proof

  40. Assignment: 5.7 P. 201 #24 – 34 (even), 42, 44 P. 179 #49 – 52

More Related